Physics, 08.12.2020 07:10, Devan6264

I need help with these questions :
(see image )

I need help with these questions :(see image )

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answered: isaacb6291

Hey There!_____________________________________Question 11)__________________________________________________________________________DATA:

Muzzle Velocity = initial velocity of bullet = V_i = 300 m/s

Height "H" dropped at horizontal distance,

a) x = 20m

b) x = 40m

c) x = 60m

d) How far will it drop after 1 second = ?

_____________________________________SOLUTION:

We have to find the Height H dropped at the respective given horizontal distances. To find H first we have to calculate the time the bullet has taken to reach at every "X"(Horizontal Distance). So,

A) X = 20m

v = 300m/s

t = ?

X = vt

20 = 300 x t

t = 0.067 seconds.

Now, Height dropped can be determine by 2nd equation of motion,

$H = V_{iy}t + \frac{1}{2}gt^2$

As initial vertical velocity is zero because the question says that the bullet is fired in the horizontal direction and there is no velocity component in y-axis or vertical axis.

$H = \frac{1}{2}gt^2$

$H = \frac{1}{2} x 10 x (0.067)^2\\\\H = (5)x(4.489x10^{-3})\\\\H = 0.022445 meters$

_____________________________________

B) X = 40m

v = 300m/s

t = ?

X = vt

40 = 300 x t

t = 0.133 seconds

Now, Height dropped can be determine by 2nd equation of motion,

$H = \frac{1}{2}gt^2\\\\H = \frac{1}{2}x10x(0.133)^2\\\\H = 0.088445 meters$

_____________________________________

C) X = 60m

v = 300m/s

t = ?

X = vt

60 = 300 x t

t = 0.2 seconds

Now, Height dropped can be determine by 2nd equation of motion,

$H = \frac{1}{2}gt^2\\\\H = \frac{1}{2}x10x(0.2)^2 \\\\H = 0.2 meters$

_____________________________________

D) By the second equation of motion,

$H=\frac{1}{2}gt^2 \\\\H = \frac{1}{2}x10x(1)^2 \\\\H = 5 meters$

______________________________________Question 12:DATA:

Angle = θ(theta) = 30

Initial velocity = V_0 = 200m/s

Time to reach the ground = T = ?

Range = R = ?

_____________________________________SOLUTION:

The total time of flight is given by,

$T = \frac{2V_0Sin(theta)}{g}\\\\T = \frac{2(200)Sin(30)}{10} \\\\T = \frac{200}{10}\\\\T = 20 seconds$

the Range of projectile is given by,

$R = \frac{V_0^2}{g}Sin2(theta)\\\\R = \frac{200^2}{10} Sin2(30)\\\\R = \frac{40000}{10}Sin60\\\\R = (4000)x(0.866)\\\\R = 3464.1 meters$

_____________________________________Best Regards,'Borz'

answered: Guest

Ithink your answer is d

answered: Guest

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way

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